dispatch_semaphore_t重用 - 我在这里缺少什么？

Question

我有一些代码,我使用dispatch_semaphore_t来指示操作完成.当信号量是成员变量时,它似乎行为不正确.我将展示有效的示例代码和一个似乎不起作用的示例:

@implementation someClass  
{  
  dispatch_semaphore_t memberSem;  
  dispatch_semaphore_t* semPtr;  
  NSThread* worker;
  BOOL taskDone;  
}  

- (id)init  
{  
  // Set up the worker thread and launch it - not shown here.  
  memberSem= dispatch_semaphore_create(0); 
  semPtr= NULL;  
  taskDone= FALSE;  
}  

- (void)dealloc  
{  
  // Clean up the worker thread as needed - not shown here.  
  if((NULL != semPtr) && (NULL != *semPtr))  
    disptatch_release(*semPtr);  

  dispatch_release(memberSem);  
}  

- (void)doSomethingArduous  
{  
  while([self notDone])  // Does something like check a limit.  
    [self doIt];  // Does something like process data and increment a counter.  

  taskDone= TRUE;  // I know this should be protected, but keeping the example simple for now.  

  if((NULL != semPtr) && (NULL != *semPtr))  
    dispatch_semaphore_signal(*semPtr);  // I will put a breakpoint here, call it  "SIGNAL"  
}  

- (BOOL)getSomethingDoneUseLocalSemaphore  
{  
  taskDone= FALSE;  // I know this should be protected, but keeping the example simple for now.  
  dispatch_semaphore_t localSem= dispatch_semaphore_create(0);  
  semPtr= &localSem;  
  [self performSelector:doSomethingArduous onThread:worker withObject:nil waitUntilDone:NO];  

  dispatch_time_t timeUp= dispatch_time(DISPATCH_TIME_NOW, (uint64_t)(2.5 * NSEC_PER_SEC));  
  dispatch_semaphore_wait(localSem, timeUp);  

  semPtr= NULL;  
  dispatch_release(localSem);  

  // I know I could just return taskDone. The example is this way to show what the problem is.  
  if(taskDone)  // Again with thread safety.  
    return TRUE;    

  return FALSE;  
}  

- (BOOL)getSomethingDoneUseMemberSemaphore  
{  
  taskDone= FALSE;  // I know this should be protected, but keeping the example simple for now.  

  semPtr= &memberSem;  // I will put a breakpoint here, call it "START"  
  [self performSelector:doSomethingArduous onThread:worker withObject:nil waitUntilDone:NO];  

  dispatch_time_t timeUp= dispatch_time(DISPATCH_TIME_NOW, (uint64_t)(2.5 * NSEC_PER_SEC));  
  dispatch_semaphore_wait(memberSem, timeUp);  

  semPtr= NULL;  

  // I know I could just return taskDone. The example is this way to show what the problem is.  
  if(taskDone)  // Again with thread safety.  
    return TRUE;  // I will put a breakpoint here, call it "TASK_DONE"   

  return FALSE;  // I will put a breakpoint here, call it "TASK_NOT_DONE"  
}  

- (void)hereIsWhereWeBringItTogether  
{  
  BOOL gotItDoneLocal= [self getSomethingDoneUseLocalSemaphore];  // Will return TRUE.  
  gotItDoneLocal= [self getSomethingDoneUseLocalSemaphore];  // Will return TRUE.  
  gotItDoneLocal= [self getSomethingDoneUseLocalSemaphore];  // Will return TRUE.  

  BOOL gotItDoneMember= [self getSomethingDoneUseMemberSemaphore];  // Will return TRUE. I will put a breakpoint here, call it "RUN_TEST"  
  gotItDoneMember= [self getSomethingDoneUseMemberSemaphore];  // Will return FALSE.  
}  

所以,考虑到代码和我得到/得到的结果,我把断点按照我的实际代码中的描述:一个在main函数中,一个在工作函数中开始,一个在成员信号量发出信号,两个在等待.

我发现的是在我使用成员信号量的情况下,在第一轮我停在断点"RUN_TEST",运行并点击断点"START",然后运行命中断点"SIGNAL",然后运行命中断点"TASK_DONE" - 一切如预期.

当我继续运行时,我点击断点"START",然后运行命中断点"TASK_NOT_DONE",然后运行命中断点"SIGNAL"

也就是说,当我使用作为成员的信号量运行序列,并执行看起来像正确的信号/等待时,我第二次尝试等待该信号量时,我似乎在吹嘘并且在我退出之后它会发出信号等待.

我似乎要么不管理计数权(信号/等待配对),要么成员信号量不会回到无信号状态.

我的感觉是,我在这里缺少一些基本的东西.任何输入将不胜感激.

编辑:最终我似乎缺少的是由于我的实际代码有点复杂.而不是从艰巨的任务中获得干净的回报,而是涉及多个线程和一个postNotification.我用通知处理程序中的代码替换了postNotification - 它设置了一个标志并用信号通知信号量.这样就消除了通知处理程序可能引入的任何延迟.

Answer 1

是的,这是预期的行为.如果你等待一个信号超时,当信号到来时,它将被下一个dispatch_semaphore_wait特定信号量的调用捕获.请考虑以下示例:

例如:

dispatch_semaphore_t semaphore = dispatch_semaphore_create(0);
dispatch_time_t timeout;

// in 5 seconds, issue signal

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
    sleep(5);
    NSLog(@"Signal 1");
    dispatch_semaphore_signal(semaphore);
});

// wait four seconds for signal (i.e. we're going to time out before the signal)

NSLog(@"Waiting 1");
timeout = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(4.0 * NSEC_PER_SEC));
if (dispatch_semaphore_wait(semaphore, timeout))
    NSLog(@"Waiting for 1: timed out");
else
    NSLog(@"Waiting for 1: caught signal");

// now, let's issue a second signal in another five seconds

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
    sleep(5);
    NSLog(@"Signal 2");
    dispatch_semaphore_signal(semaphore);
});

// wait another four seconds for signal

// this time we're not going to time out waiting for the second signal, 
// because we'll actually catch that first signal, "signal 1")

NSLog(@"Waiting 2");
timeout = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(4.0 * NSEC_PER_SEC));
if (dispatch_semaphore_wait(semaphore, timeout))
    NSLog(@"Waiting for 2: timed out");
else
    NSLog(@"Waiting for 2: caught signal");

// note, "signal 2" is still forthcoming and the above code's 
// signals and waits are unbalanced 

因此,当你使用类实例变量时,你的getSomethingDoneUseMemberSemaphore行为与上面类似,第二次调用dispatch_semaphore_wait将捕获发出的第一个信号,因为(a)它是相同的信号量; (b)如果第一次打电话dispatch_semaphore_signal超时.

但是如果你每次都使用唯一的信号量,那么第二次调用dispatch_semaphore_wait将不会响应dispatch_semaphore_signal第一个信号量.